Reinforced Concrete Suspended Floors and beam design

Ribbed Floors are used to….Ribbed Floors are used to…. reduce the overall depthreduce the overall depth of a traditional of a traditional

cast insitu reinforced concrete beam and cast insitu reinforced concrete beam and slab suspended floorslab suspended floor

The basic concept is to replace the wide spaced deep The basic concept is to replace the wide spaced deep

beams with narrow spaced shallow beams or ribs beams with narrow spaced shallow beams or ribs which will carry only a small amount of slab which will carry only a small amount of slab

loadingloading.

These floors can be designed as one or two way These floors can be designed as one or two way spanning floors.spanning floors.

One way spanning ribbed floors are sometimes One way spanning ribbed floors are sometimes called trough floorscalled trough floors

the two way spanning ribbed floors are called the two way spanning ribbed floors are called coffered or waffle floors.coffered or waffle floors.

•Ribbed floors are usually cast against metal, glass fibre or polypropylene

They are made in preformed moulds which are temporarily supported on plywood decking joists and props

From the following diagram make sure that you note down the joist data then work out the total load per joist

Answer:

Total load (W) per joist = 5m X 0.4m X 2.25 kN/m2 = 45 kN

On the following page you should

•Make a note of the bending moment (BM) formula

•Make a note of the data

•Find the depth d, of the joist

Answer

The formula for the Bending moment of a beam is ….

Transpose the BM formula to find, d.

Answer from standard sizes

TN4 Work slowly through the following

•Joist and Beam Sizing for calculating overall dimensions alone is insufficient.

• checks should also be made to satisfy: resistance to deflection•adequate safe bearing and •resistance to shear.

•Deflection — should be minimal to prevent damage to….• plastered ceilings. •An allowance of up to 0003 X span is normally acceptable; for the last example this will be:- 0003 X 5000mm = 15mm

(Use b and d from before) The formula for calculating deflection due to a uniformly distributed load is:

Answer

Answer

TN4

Use the nearest commercial size to find I then find ,d.

Answer

Answer

Important take note

The formula is

The formula is

Answer in cmin cm

TN3

I= 9504 from tables of values

Permissible deflection is 1/360 of 4 m = 11.1 cm.

Therefore actual deflection of 0.835 cm is acceptable.

Ref. BS 5950: Structural use of steelwork in building.

Structural Steelwork—Column Design

•Steel columns or stanchions have a tendency to….

•buckle or bend under extreme loading. This can be attributed to:

(a) length

(b) cross sectional area

(c) method of end fixing

(d) the shape of section

(b) and (d) are incorporated into a geometric property of section, known as the radius of gyration (r).

It can be calculated:- r = SQRT (1/A)

where: I = (Inertia) -2nd moment of area

A = cross sectional area

Note: r,I and A are all listed in steel design tables, e.g.. BS4:1980.

Answer

Answer

Answer

TN3